Name:_______________________ ___
H-ITT or pseudonym: ______________
PHY2061
10-21-04
Exam 2 Solutions
R1
a
+
ε
R2
R3
b
1. Consider the circuit above. The battery supplies an EMF of ε = 6.0 V, and the resistances
of the 3 resistors are R1 = 100Ω , R2 = 25Ω , and R3 = 75Ω
(a) [6 points] Find the voltage drop between points a and b: ∆Vab = Va − Vb .
Let’s first find the equivalent resistance of R2 and R3:
1
1
1
1
1
=
+
=
+
Reff R2 R3 25 75
⇒ Reff = 18.75Ω
The total resistance of the circuit is thus Rtot = R1 + Reff = 118.75Ω
The current in the circuit, by Ohm’s Law, is:
i=
ε
= 0.0505A
Rtot
So the voltage drop is
∆Vab = iReff = ( 0.0505 )(18.75 ) = 0.95 V
(b) [6 points] Find the current through resistor R2.
Knowing the voltage drop ∆Vab , the current i2 is found by:
∆V
0.95 V
i2 = ab =
= 0.038 A
25Ω
R2
(c) [6 points] Find the total power dissipated in the circuit that must be provided by the
battery.
P = i 2 Rtot = ( 0.0505 A ) (118.75Ω ) = 0.303 Watts
2
Page 1 of 9
Name:_______________________ ___
H-ITT or pseudonym: ______________
PHY2061
10-21-04
(d) [8 points] Suppose an identical circuit (with the same resistors and battery
configuration) is connected to the above circuit at points a and b. That is, a conductor
connects point a of the first circuit to point a of the second circuit, and another
conductor connects point b of the first circuit to point b of the second. Calculate the
voltage drop between points a and b for this combined circuit. (Hint: apply Kirchoff’s
laws to each loop of the circuit.)
R1
i1
ε
a
R1
a
i3
+
R2
i2
+
R2
R3 R3
ε
b
b
First find the effective resistance of all 4 middle resistors in parallel:
1
2
2
2
2
=
+
=
+
Reff R2 R3 25 75
⇒ Reff = 9.375Ω
Let i1 be the current leaving the left battery, i2 that leaving the second, and i3 being
that traveling down the center branch. By Kirchoff’s First Law:
i3 = i1 + i2
By Kirchoff’s Second Law applied to each loop:
i1 R1 + i3 Reff − ε = 0 ⇒ i1 R1 + i1 Reff + i2 Reff − ε = 0
i2 R1 + i3 Reff − ε = 0 ⇒ i2 R1 + i1 Reff + i2 Reff − ε = 0
These two equations can only be true if i1 = i2 , as you might expect from symmetry.
Solving for i1 gives:
i1 R1 + 2i1 Reff − ε = 0
i1 =
ε
R1 + 2 Reff
Now i3 = 2i1
2ε Reff
= 0.95 V as before
R1 + 2 Reff
(Exactly as you would expect connecting two batteries of same voltage in parallel.)
⇒ ∆Vab = i3 Reff =
Page 2 of 9
Name:_______________________ ___
H-ITT or pseudonym: ______________
PHY2061
10-21-04
R1
a
+
ε
C1
C2
S
b
3. Consider the circuit above. A battery supplies an EMF of ε = 12.0 V, the resistance
R1 = 1.25 MΩ , and the capacitors have capacitance C1 = 18µ F and C1 = 9µ F . Consider that
switch S is closed at time t = 0. ( 1 MΩ = 106 Ω , 1µ F = 10-6 F ).
(a) [6 points] What is the time constant of the circuit that governs how fast the current
changes after the switch is closed ?
The RC time constant governs how fast the current changes and the capacitors charge.
The effective capacitance of two capacitors in series is:
1
1
1
= +
⇒ Ceff = 6 µ F
Ceff C1 C2
τ C = RCeff = 1.25 × 106 × 6 × 10−6 = 7.5 s
(b) [6 points] What is the maximum charge attained on the top plate of capacitor C1
(i.e. at point a) after the switch is closed?
The maximum voltage drop across the pair of capacitors will be when the current
through the circuit stops, which implies 0 voltage drop across the resistor. Thus,
∆Vab ( t → ∞ ) = ε = 12V =
qmax
Ceff
qmax = (12V ) ( 6 × 10−6 F ) = 7.2 ×10−5 C
The same |q| is stored on each capacitor plate, and it is that for the equivalent
capacitance of 2 capacitors in series.
Page 3 of 9
PHY2061
10-21-04
Name:_______________________ ___
H-ITT or pseudonym: ______________
(c) [6 points] Sketch the voltage drop ∆Vab = Va − Vb across the capacitors as a function of
time t. Try to be accurate in your sketch.
Page 4 of 9
Name:_______________________ ___
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PHY2061
10-21-04
3. A new type of cannon can launch 5 kg projectiles at the high velocity of 0.3c
(a) [6 points] What is the minimum amount of energy required to accelerate a projectile
from rest to this final velocity?
The energy required is the work done to change the kinetic energy of the mass:
W = K = ( γ − 1) mc 2
γ=
1
1 − v2 / c2
= 1.0483
W = 0.0483 ( 5 kg ) ( 3 × 108 m/s ) = 2.17 ×1016 J
2
(b) [6 points] If the cannon is mounted on the front of a spaceship traveling away from
Earth at a velocity of 0.6c, what will be the velocity of the launched projectile in the
reference frame of the Earth?
Use the Lorentz transformation of velocity:
ux′ + v
ux =
1 + u x′ v / c 2
ux =
0.3c + 0.6c
= 0.763c = 2.29 × 108 m / s
1 + 0.3 0.6
a fa f
Page 5 of 9
Name:_______________________ ___
H-ITT or pseudonym: ______________
PHY2061
10-21-04
4. A kaon is an unstable particle with a rest mass energy of 500 MeV and an average
lifetime of τ = 5.2 ×10−8 s in its rest frame. Consider a beam of kaons, where each kaon
has a momentum of 1000 MeV/c as measured in a laboratory. (1 MeV = 1.6 ×10−13 J )
(a) [6 points] What is the velocity of each kaon in the laboratory?
p = γ mv = γ mc
pc = γ mc 2
v
c
v
c
v2
2
p 2c 2 = m2c 4 c 2
v
1− 2
c
2
2 2
⎛ v ⎞pc
v2
⇒ ⎜1 − 2 ⎟ 2 4 = 2
c
⎝ c ⎠m c
⇒
v
pc
=
c mc 2
1
p 2c2
m2c 4
⇒ v = 0.894c = 2.68 × 108 m/s
1+
(b) [6 points] What is the average distance the kaons travel at this momentum in the
laboratory before decaying?
γ=
1
2
v
1− 2
c
=
1
1 − 0.8942
= 2.236
d = vt = γ vτ = ( 2.236 )( 0.894 ) ( 3 × 108 )( 5.2 × 10−8 ) = 31.2 m
Page 6 of 9
Name:_______________________ ___
H-ITT or pseudonym: ______________
PHY2061
10-21-04
5. A long straight wire has an electric charge density of λ = 7.0µ C / m along its length,
where 1µ C = 10−6 C , as measured in the rest frame of the wire.
(a) [6 points] What is the observed charge density of the wire as measured by an
observer traveling at a velocity of v = 2.4 ×108 m/s parallel to the direction of the wire?
Length is contracted along the direction of motion: L =
L0
γ
, but the electric charge is
invariant in some region. So the charge density per unit length increases:
λ=
γ=
q
q
=γ
= γλ0
L
L0
1
=
1
=
25
= 1.667
9
1 − 0.82
v2
c2
⇒ λ = 1.667 ( 7 µ C / m ) = 11.67 µC / m
1−
(b) [6 points] What is the observed charge density of the wire as measured by an
observer traveling at a velocity of v = 2.4 ×108 m/s perpendicular to the direction of the
wire?
Dimensions perpendicular to the direction of travel are unaffected by Lorentz
Transformations, so the charge density remains the same.
Page 7 of 9
Name:_______________________ ___
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PHY2061
10-21-04
y
−
−
− −
−
−
E
x
e+ + + + + + +
6. [8 points] A beam of electrons (“cathode rays”) is sent between two parallel electric plates
separated by a distance d = 1.5 cm with a potential difference of 300 V between them. The
electric field points in the + yˆ direction. If the electron beam travels perpendicular to the
electric field in the +xˆ direction, and each electron has a kinetic energy of 5,000 eV, what
magnetic field is necessary (direction and magnitude) so that the electrons continue traveling
in a straight line without deflection by the electric field? The charge of the electron is
q = −e = −1.6 × 10−19 C , the electron rest-mass energy is 511,000 eV, and
1 eV = 1.6 ×10−19 J .
Balance forces:
F = q (E + v × B ) = 0
⇒ E+ v×B = 0
E
zˆ
v
300 V
V
E= =
= 2 ×104 N/C
d 0.015 m
1
K = mv 2 (check that it is non-relativistic!)
2
⇒B=
⇒v=c
⇒B=
2 ( 5000 eV )
2K
8
3
10
m/s
=
×
= 0.14c = 4.2 × 107 m/s (yes, it's non-rel.)
(
)
2
511, 000 eV
mc
V 2 × 104 N/C
=
= 0.48 mT (0.48 Gauss)
4.2 × 107
dv
Page 8 of 9
Name:_______________________ ___
H-ITT or pseudonym: ______________
PHY2061
10-21-04
iA
d
C
iB
7. Consider two parallel wires, A and B, separated by a distance d = 0.5 cm as shown above.
The current of wire A is iA = 9 A , and the current of wire B is iB = 3 A , both in the
x̂ direction.
(a) [6 points] Calculate the direction and magnitude of the magnetic field at point C,
midway between the two wires in the plane of the wires.
Each wire generates a magnetic field:
2ki
B wire =
rˆ
r
At point C, the field from A points into the plane of the paper, and the field from B
out of the plane. The field from A in stronger, so the net field points into the plane of
the paper:
B = B A + BB
| B |=
2k
( iA − iB )
(d / 2)
( 4 ×10
=
−7
T ⋅ m /A ) ( 9 − 3 A )
0.005 m
= 4.8 ×10−4 T
(b) [6 points] Find the direction and magnitude of the force per unit length (per meter)
on wire B arising from the interaction of its current with the magnetic field of wire A.
F = i L×B
The force is attractive between the two wires with currents in the same direction. Per
unit length, the force is:
2kiAiB ( 2 × 10
=
F = iB BA =
d
F = 1.08 × 10−3 N
−7
T ⋅ m / A ) ( 9 A )( 3 A )
0.005 m
Page 9 of 9